Question Number 56612 by necx1 last updated on 19/Mar/19
Answered by mr W last updated on 19/Mar/19
$${a}_{\mathrm{1}} ={acceleration}\:{of}\:{small}\:{block} \\ $$$${a}_{\mathrm{2}} ={acceleration}\:{of}\:{big}\:{block} \\ $$$${f}={friction}\:{between}\:{both}\:{blocks} \\ $$$$ \\ $$$${assume}\:{a}_{\mathrm{1}} ={a}_{\mathrm{2}} ={a}: \\ $$$${ma}_{\mathrm{1}} ={f}\leqslant\mu{mg}=\mathrm{0}.\mathrm{5}{mg} \\ $$$${mg}−{f}=\mathrm{2}{ma}_{\mathrm{2}} \\ $$$${mg}−{ma}_{\mathrm{1}} =\mathrm{2}{ma}_{\mathrm{2}} \\ $$$${g}−{a}=\mathrm{2}{a} \\ $$$$\Rightarrow{a}=\frac{{g}}{\mathrm{3}} \\ $$$${f}={ma}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}{mg}<\mathrm{0}.\mathrm{5}{mg} \\ $$$${i}.{e}.\:{assumption}\:{is}\:{correct}.\:{both}\:{blocks} \\ $$$${move}\:{the}\:{same}\:{distance}\:{d}. \\ $$$${work}\:{done}\:{by}\:{force}\:{f}\:{is} \\ $$$${W}={fd}=\frac{{mgd}}{\mathrm{3}} \\ $$$$\Rightarrow{answer}\:\left(\mathrm{1}\right) \\ $$
Commented by necx1 last updated on 20/Mar/19
$${Thank}\:{you}\:{so}\:{much}\:{sir}.{I}\:{really} \\ $$$${understand}\:{this}\:{approach}. \\ $$
Commented by mr W last updated on 20/Mar/19
$${Can}\:{you}\:{solve}\:{the}\:{same}\:{question}\:{if} \\ $$$${a}\:{force}\:{of}\:\mathrm{2}{mg}\:{is}\:{applied}\:{on}\:{the}\:{big} \\ $$$${block}\:{and}\:{the}\:{big}\:{block}\:{moves}\:{a}\:{distance} \\ $$$${of}\:{d}? \\ $$$${assume}\:{the}\:{big}\:{block}\:{is}\:{big}\:{enough}\:{such} \\ $$$${that}\:{the}\:{small}\:{block}\:{always}\:{remains} \\ $$$${on}\:{the}\:{big}\:{block}. \\ $$
Commented by necx1 last updated on 20/Mar/19
$${ok}\:{sir}.....\:{I}'{ll}\:{do}\:{that}.{I}'{m}\:{in}\:{a}\:{class} \\ $$$${presently}. \\ $$
Commented by mr W last updated on 22/Mar/19
$${if}\:{you}\:{get}\:{the}\:{same}\:{result},\:{don}'{t}\:{wonder}, \\ $$$${it}'{s}\:{just}\:{an}\:{accident}. \\ $$